If a1,a2,a3,…,a24 are  in arithmetic progression  and  a1+a5+a10+a15+a20+a24=225 then a1+a2+a3+…+a23+a24 is equal to

If a1,a2,a3,,a24 are  in arithmetic progression  and  a1+a5+a10+a15+a20+a24=225 then a1+a2+a3++a23+a24 is equal to

  1. A

    909

  2. B

    75

  3. C

    750

  4. D

    900

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    Solution:

    Given that,

      a1+a5+a10+a15+a20+a24=225

        a1+a24+a5+a20+a10+a15=225    3a1+a24=225    a1+a24=75

    [in an AP the sum of the terms equidistant from the beginning and the end is same and is equal to the sum of the first and last term]

      a1+a2++a24=n2(a+l)=242a1+a24=12×75=900  [from eq.(i)]

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